A certain problem that I have been working on involves the equation

$$1 = \frac{1}{1-n}$$

One can see that the only real-number solution is $n=0$. As far as the original problem goes, that is enough. However, I have been thinking about extending the problem into the complex numbers. I want to find all complex numbers $z$ that satisfy

$$\left|\frac{1}{1-z}\right| = 1$$

In my attempt to solve the problem, I rewrite the equation as

$$\left| \frac{-1}{z-1} \right| = 1 $$

My intuition is that the magnitude of a quotient of complex numbers is the same as the quotient of their magnitudes, though I can not see a way to prove it. If that is true, then -1 has a magnitude of 1, and I must find the set of complex numbers where $|z-1|$ is 1.

That set can be visualized in the complex plane as the unit circle translated one unit to the right. Can someone check whether my assumption is true and answer is correct?

Thank you.


1 Answer 1


You are spot on with your assumption. In fact, division in the complex plane can be defined as $$\frac{z_1}{z_2}=\frac{z_1\bar{z_2}}{|z_2|^2}$$ where $\bar{z_2}$ is the complex conjugate of $z_2$.

In terms of the set of solutions for your problem, it is as simple as $|z-1|=1$ which can be written $$r(\cos(\theta)+1+i\sin(\theta))$$ Among many other ways.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .